A wholesaler of nuts sells two mixtures of peanuts and cashews. The first mixture is 3/4 peanuts by weight. The...

1. Translate the problem requirements: We have two mixtures with the same amount of cashews. The first is \(\frac{3}{4}\) peanuts, the second has 8 fewer pounds of peanuts and is \(\frac{7}{10}\) peanuts. We need to find the total weight of the first mixture.
2. Set up variables for the first mixture: Define the total weight of the first mixture and express its peanut and cashew components using the \(\frac{3}{4}\) ratio.
3. Express the second mixture in terms of the first: Use the fact that cashews remain the same while peanuts decrease by 8 pounds, then calculate the new total weight.
4. Apply the second mixture's ratio constraint: Set up an equation using the fact that peanuts are \(\frac{7}{10}\) of the second mixture's weight.
5. Solve for the first mixture's total weight: Solve the equation to find the answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in simple terms:

We have two mixtures of nuts - both contain peanuts and cashews.

For the first mixture:
• \(\frac{3}{4}\) of its weight is peanuts
• The remaining \(\frac{1}{4}\) of its weight is cashews

For the second mixture:
• It has the same amount of cashews as the first mixture (same pounds of cashews)
• It has 8 fewer pounds of peanuts than the first mixture
• \(\frac{7}{10}\) of its weight is peanuts

Our goal: Find the total weight of the first mixture.

Process Skill: TRANSLATE - Converting the word problem into clear mathematical relationships

2. Set up variables for the first mixture

Let's call the total weight of the first mixture "W" pounds.

Since the first mixture is \(\frac{3}{4}\) peanuts:
• Peanuts in first mixture = \(\frac{3}{4} \times \mathrm{W}\) pounds
• Cashews in first mixture = \(\frac{1}{4} \times \mathrm{W}\) pounds

We can check this makes sense: \(\frac{3}{4}\mathrm{W} + \frac{1}{4}\mathrm{W} = \mathrm{W}\) ✓

3. Express the second mixture in terms of the first

Now let's figure out what's in the second mixture:

• Cashews in second mixture = same as first = \(\frac{1}{4} \times \mathrm{W}\) pounds
• Peanuts in second mixture = 8 fewer than first = \(\frac{3}{4} \times \mathrm{W} - 8\) pounds

So the total weight of the second mixture is:
Cashews + Peanuts = \(\frac{1}{4}\mathrm{W} + [\frac{3}{4}\mathrm{W} - 8]\)
= \(\frac{1}{4}\mathrm{W} + \frac{3}{4}\mathrm{W} - 8\)
= \(\mathrm{W} - 8\) pounds

4. Apply the second mixture's ratio constraint

Here's the key insight: we know that \(\frac{7}{10}\) of the second mixture's weight is peanuts.

The second mixture weighs \((\mathrm{W} - 8)\) pounds total.
So peanuts should equal \(\frac{7}{10} \times (\mathrm{W} - 8)\) pounds.

But we also know from step 3 that peanuts in the second mixture = \(\frac{3}{4}\mathrm{W} - 8\) pounds.

These must be equal:
\(\frac{3}{4}\mathrm{W} - 8 = \frac{7}{10} \times (\mathrm{W} - 8)\)

Process Skill: APPLY CONSTRAINTS - Using the given ratio to create our solving equation

5. Solve for the first mixture's total weight

Let's solve: \(\frac{3}{4}\mathrm{W} - 8 = \frac{7}{10}(\mathrm{W} - 8)\)

Expand the right side:
\(\frac{3}{4}\mathrm{W} - 8 = \frac{7}{10}\mathrm{W} - \frac{7}{10} \times 8\)
\(\frac{3}{4}\mathrm{W} - 8 = \frac{7}{10}\mathrm{W} - \frac{56}{10}\)
\(\frac{3}{4}\mathrm{W} - 8 = \frac{7}{10}\mathrm{W} - 5.6\)

Collect W terms on one side:
\(\frac{3}{4}\mathrm{W} - \frac{7}{10}\mathrm{W} = 8 - 5.6\)
\(\frac{3}{4}\mathrm{W} - \frac{7}{10}\mathrm{W} = 2.4\)

To subtract these fractions, let's use common denominator 20:
\(\frac{15}{20}\mathrm{W} - \frac{14}{20}\mathrm{W} = 2.4\)
\(\frac{1}{20}\mathrm{W} = 2.4\)
\(\mathrm{W} = 2.4 \times 20 = 48\)

Let's verify: If W = 48, then:
• First mixture has 36 lbs peanuts, 12 lbs cashews
• Second mixture has 28 lbs peanuts, 12 lbs cashews (total 40 lbs)
• Check: \(\frac{28}{40} = \frac{7}{10}\) ✓

Final Answer

The first mixture contains 48 pounds of nuts total.

This matches answer choice C.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "same number of pounds of cashews"
Students might incorrectly assume that the second mixture has the same proportion of cashews as the first mixture, rather than understanding that it has the exact same weight of cashews. This leads to setting up incorrect equations based on ratios instead of absolute weights.

2. Confusion about what "8 fewer pounds of peanuts" means
Some students might misinterpret this as the second mixture having 8 fewer pounds total weight rather than specifically 8 fewer pounds of peanuts. This error leads to incorrect setup of the second mixture's composition.

3. Missing the connection between absolute weights and ratios
Students may struggle to connect that while the first mixture uses fractional descriptions (\(\frac{3}{4}\) peanuts), they need to work with absolute weights to handle the "8 fewer pounds" constraint. This can lead to purely ratio-based approaches that miss the absolute weight relationships.

Errors while executing the approach

1. Arithmetic errors when finding common denominators
When solving \(\frac{3}{4}\mathrm{W} - \frac{7}{10}\mathrm{W} = 2.4\), students commonly make mistakes converting to the common denominator of 20, incorrectly calculating \(\frac{3}{4}\) as \(\frac{12}{20}\) instead of \(\frac{15}{20}\), or \(\frac{7}{10}\) as \(\frac{16}{20}\) instead of \(\frac{14}{20}\).

2. Sign errors when rearranging the constraint equation
When moving from \(\frac{3}{4}\mathrm{W} - 8 = \frac{7}{10}(\mathrm{W} - 8)\) to collect terms, students often make sign errors, particularly when distributing the \(\frac{7}{10}\) or when moving terms across the equals sign, leading to incorrect coefficients.

3. Decimal conversion errors
Students might incorrectly convert \(\frac{56}{10}\) to something other than 5.6, or make errors in the final multiplication \(2.4 \times 20\), especially if they try to work entirely in fractions and make conversion mistakes.

Errors while selecting the answer

No likely faltering points - the final calculation clearly yields 48, which directly matches answer choice C, making selection errors unlikely.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a strategic value for the first mixture
Since the first mixture is \(\frac{3}{4}\) peanuts by weight, let's choose a total weight that makes fraction calculations clean. Let's try 80 pounds for the first mixture (divisible by 4, making \(\frac{3}{4}\) calculations easy).

Step 2: Calculate components of the first mixture
First mixture total weight = 80 pounds
Peanuts in first mixture = \(\frac{3}{4} \times 80 = 60\) pounds
Cashews in first mixture = \(\frac{1}{4} \times 80 = 20\) pounds

Step 3: Determine the second mixture components
The second mixture has the same cashews but 8 fewer peanuts:
Cashews in second mixture = 20 pounds (same as first)
Peanuts in second mixture = 60 - 8 = 52 pounds
Total weight of second mixture = 20 + 52 = 72 pounds

Step 4: Verify using the second mixture's ratio
According to the problem, peanuts should be \(\frac{7}{10}\) of the second mixture:
Expected peanuts = \(\frac{7}{10} \times 72 = 50.4\) pounds
Our calculated peanuts = 52 pounds

Step 5: Adjust and find the correct value
Since 52 ≠ 50.4, our guess of 80 wasn't correct. However, this confirms the approach works. The actual answer from the algebraic solution is 48 pounds.

Let's verify with 48:
First mixture: 48 pounds total
Peanuts = \(\frac{3}{4} \times 48 = 36\) pounds
Cashews = \(\frac{1}{4} \times 48 = 12\) pounds

Second mixture:
Cashews = 12 pounds (same)
Peanuts = 36 - 8 = 28 pounds
Total = 12 + 28 = 40 pounds

Check: \(\frac{7}{10} \times 40 = 28\) pounds of peanuts ✓

This confirms that 48 pounds is the correct answer.